Let E, F be points on the side AB of a triangle ABC. Prove that if ∠ECA = ∠BCF, then |AE||AF| / |BE||BF| = |AC|^2/|BC|^2 
Let E, F be points on the side AB of a triangle ABC. Prove
  that if $∠ECA = ∠BCF$, then
  $${|AE||AF|\over |BE||BF|}={|AC|^2\over |BC|^2}$$

Please solve using power of a point.
I have been trying to solve this problem for a while and this is what I have so far:
We must consider the circle through points C, E, and F. We can then obtain the following equations by using the power of a point with respect to a circle:
Let G be the intersection point of the side AC of the triangle with the circle containing C, E, and F. Let H be the intersection point of the side BC of the triangle with the circle containing C, E, and F.


*

*|AE||AF|= |AG||AC| by power of a point

*|BF||BE|= |BH||BC| by power of a point


We also have that the length of the arc from G to E of the circle equals the length of the arc from F to H because the angles subtended by these arcs are the same as stated in the problem. This also implies that the lengths of the chords of these arcs are also the same.
I am not sure where to go from here. It seems that I should try to prove that |AG|/|BH| = |AC|/|BC|. Possibly with similar triangles?
Any help would be much appreciated.
 A: 
Let $\theta=\widehat{ACE}=\widehat{FCB}$ and $\varphi=\widehat{ACF}=\widehat{ECB}$. By the sine theorem applied to $ACE$ and $FCB$ we have
$$ \frac{AE}{FB}=\frac{AE/\sin\theta}{FB/\sin\theta}=\frac{AC/\sin\widehat{AEC}}{BC/\sin\widehat{BFC}}=\frac{AC/\sin\widehat{BEC}}{BC/\sin\widehat{AFC}}$$
and by applying the sine theorem to $BEC$ and $AFC$ the last ratio equals
$$ \frac{AC}{BC}\cdot\frac{\sin\widehat{AFC}}{\sin\widehat{BEC}}=\frac{AC^2}{BC^2}\cdot\frac{\sin\widehat{AFC}/AC}{\sin\widehat{BEC}/BC}=\frac{AC^2}{BC^2}\cdot\frac{\sin\varphi/AF}{\sin\varphi/BE}$$
hence
$$ \frac{AE\cdot AF}{BE\cdot BF}=\frac{AC^2}{BC^2} $$
as wanted. This identity can be stated as "there is a circle orthogonal to both $\Gamma_{CEF}$ and $\Gamma_{CAB}$", which is fairly trivial since $\Gamma_{CEF}$ and $\Gamma_{CAB}$ are tangent at $C$.
A: 
As shown in the diagram, we first reflect the figure about the angle bisector of $\angle ACB$ and then invert the figure about a circle centred at $C$ with radius $r=\sqrt{BC.AC}
$. The net effect of the transformation is that it maps $A$ to $B$ and vice versa. The line AB is mapped to the circumcircle of $ABC$. Thus, $F$ is mapped to the intersection of $CE$ with the circumcircle of $ABC$ and likewise for $E$.
Reflection preserves distances and inversion gives distances according to the inversion distance formula.
Therefore,
$$B''F''=\frac{r^2}{CB.CF}.BF,\  B''E''=\frac{r^2}{CB.CE}.BE,\  A''F''=\frac{r^2}{CA.CF}.AF\text{ and } A''E''=\frac{r^2}{CA.CE}.AE \text{ .}$$
But $B''F''=A''E''$ and $B''E''=A''F''$ (as they subtend the same angle on the circumcircle), whence $$\frac{AE\cdot AF}{BE\cdot BF}=\frac{AC^2}{BC^2}\ .$$
$\blacksquare$
A: Notice that $$\angle FEG = \pi- \angle GCF = \pi-\angle ECH =\pi- \angle EGH$$
which means that $GH||EF$. Now by the Thales theorem we have $${AG\over BH} = {AC\over BC}$$ and you are done.
